Question

Charges $ +q $ and $ -q $ are placed at points $ A $ and $ B $ respectively which are a distance $ 2 \,L $ apart, $ C $ is the midpoint between $ A $ and $ B $ . The work done in moving a charge $ +Q $ along the semicircle $ CRD $ is

• A. $ \frac{qQ}{4\pi\varepsilon_{0}L} $
• B. $ \frac{qQ}{2\pi\varepsilon_{0}L} $
• C. $ \frac{qQ}{6\pi\varepsilon_{0}L} $
• D. $ -\frac{qQ}{6\pi\varepsilon_{0}L} $

Answer 1

Answer: (D)

Work done is equal to change in potential energy. In $I^{st}$ case, when charge $+Q$ is situated at $C$.

Electric potential energy of system
$U_{1}=\frac{1}{4\pi\varepsilon_{0}} \frac{\left(q\right)\left(-q\right)}{2L}+\frac{1}{4\pi\varepsilon_{0}} \frac{\left(-q\right)Q}{L}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{qQ}{L}$
In $II^{nd}$ case, when charge $+Q$ is moved from $C$ to $D$.

Electric potential energy of system in that case
$U_{2}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{\left(q\right)\left(-q\right)}{2L}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{qQ}{3L}+\frac{1}{4\pi\varepsilon_{0}} \frac{\left(-q\right)\left(Q\right)}{L}$
$∴$ Work done $= ΔU = U_2 − U_1$
$=\left(-\frac{1}{4\pi\varepsilon_{0}} \frac{q^{2}}{2L}+\frac{1}{4\pi\varepsilon_{0}} \frac{qQ}{3L}-\frac{1}{4\pi\varepsilon_{0}} \frac{qQ}{L}\right)$
$-\left(-\frac{1}{4\pi\varepsilon_{0}} \frac{q^{2}}{2L}-\frac{1}{4\pi\varepsilon_{0}} \cdot\frac{qQ}{L}+\frac{1}{4\pi\varepsilon}\cdot\frac{qQ}{L}\right)$
$=\frac{qQ}{4\pi\varepsilon_{0}}\cdot\left[\frac{1}{3L}-\frac{1}{L}\right]=\frac{qQ}{4\pi\varepsilon_{0}} \frac{\left(1-3\right)}{3L}$
$=\frac{-2qQ}{12\pi\varepsilon_{0}L}=-\frac{qQ}{6\pi\varepsilon_{0}L}$